Chapter 16, Problem 49P

(a)

To determine

The magnitude of the electric filed at a distance from the sphere’s center of $0.15\text{m}$ .

Answer to Problem 49P

The electric field at point distant $0.15\text{m}$ from the center of the sphere is zero.

Explanation of Solution

Given info:

The radius of the sphere is $R=3\text{m}$ .

The charge on the sphere is $Q=-3.5\mu \text{C}$ .

Formula used:

Gauss law states that the net electric flux passing through closed surface is equal to the $\frac{1}{{\epsilon }_0}$ times the net charge enclosed within the surface.

The expression for the Gauss Law is,

  ${\varphi }_E=\frac{q_{\text{net}}}{{\epsilon }_0}$

Calculation:

As the charge resides on the surface of the sphere, any surface enclosing the sphere below the surface would not carry any charge. Hence, from Gauss Law, the electric field at point distant $0.15\text{m}$ from the center of the sphere is zero.

Conclusion:

Thus, the electric field at point distant $0.15\text{m}$ from the center of the sphere is zero.

(b)

To determine

The magnitude of the electric filed at a distance from the sphere’s center of $2.90\text{m}$ .

Answer to Problem 49P

The electric field at point distant $2.90\text{m}$ from the center of the sphere is zero.

Explanation of Solution

Given info:

The radius of the sphere is $R=3\text{m}$ .

The charge on the sphere is $Q=-3.5\mu \text{C}$ .

Formula used:

Gauss law states that the net electric flux passing through closed surface is equal to the $\frac{1}{{\epsilon }_0}$ times the net charge enclosed within the surface.

The expression for the Gauss Law is,

  ${\varphi }_E=\frac{q_{\text{net}}}{{\epsilon }_0}$

Calculation:

As the charge resides on the surface of the sphere, any surface enclosing the sphere below the surface would not carry any charge. Hence, from Gauss Law, the electric field at point distant $2.90\text{m}$ from the center of the sphere is zero.

Conclusion:

Thus, the electric field at point distant $2.90\text{m}$ from the center of the sphere is zero.

(c)

To determine

The magnitude of the electric filed at a distance from the sphere’s center of $3.10\text{m}$ .

Answer to Problem 49P

The electric field at point distant $3.10\text{m}$ from the center of the sphere is $3.27\times {10}^3\text{N}/\text{C}$ .

Explanation of Solution

Given info:

The radius of the sphere is $R=3\text{m}$ .

The charge on the sphere is $Q=-3.5\mu \text{C}$ .

Formula used:

For points lying outside the sphere, the electric filed is the same as if all the charge were concentrated at the centre as a point charge.

  $E=\frac{kQ}{r^2}$

Calculation:

Substituting the given values in formula $E=\frac{kQ}{r^2}$ , we get

  $\begin{array}{c}E=\frac{\left(8.998\times {10}^9\right)\left(3.50\times {10}^{-6}\right)}{{\left(3.10\right)}^2}\text{N}/\text{C}\\ =3.27\times {10}^3\text{N}/\text{C}\end{array}$

Conclusion:

Thus, the electric field at point distant $3.10\text{m}$ from the center of the sphere is $3.27\times {10}^3\text{N}/\text{C}$ .

(d)

To determine

The magnitude of the electric filed at a distance from the sphere’s center of $6.00\text{m}$ .

Answer to Problem 49P

The electric field at point distant $6.00\text{m}$ from the center of the sphere is $8.74\times {10}^2\text{N}/\text{C}$ .

Explanation of Solution

Given info:

The radius of the sphere is $R=3\text{m}$ .

The charge on the sphere is $Q=-3.5\mu \text{C}$ .

Formula used:

For points lying outside the sphere, the electric filed is the same as if all the charge were concentrated at the centre as a point charge.

  $E=\frac{kQ}{r^2}$

Calculation:

Substituting the given values in formula $E=\frac{kQ}{r^2}$ , we get

  $\begin{array}{c}E=\frac{\left(8.998\times {10}^9\right)\left(3.50\times {10}^{-6}\right)}{{\left(6\right)}^2}\text{N}/\text{C}\\ =8.74\times {10}^2\text{N}/\text{C}\end{array}$

Conclusion:

Thus, the electric field at point distant $6.00\text{m}$ from the center of the sphere is $8.74\times {10}^2\text{N}/\text{C}$ .

(e)

To determine

Whether the answer will differ if the sphere is a thin shell.

Answer to Problem 49P

The answer will not change.

Explanation of Solution

Given info:

The radius of the sphere is $R=3\text{m}$ .

The charge on the sphere is $Q=-3.5\mu \text{C}$ .

Formula used:

For points lying outside the sphere, the electric filed is the same as if all the charge were concentrated at the centre as a point charge.

  $E=\frac{kQ}{r^2}$

Calculation:

The answers would not change even if it were a thin metal sheet.

Conclusion:

Thus, the answer will not change.